A proof of topological completeness for S4 in (,) Grigori Mints and Ting Zhang 2 Philosophy Department, Stanford University mints@csli.stanford.edu 2 Computer Science Department, Stanford University tingz@cs.stanford.edu Summary. The completeness of the modal logic S4 for all topological spaces as ell as for the real line R, the n-dimensional Euclidean space R n and the segment (, ) etc. (ith interpreted as interior) as proved by McKinsey and Tarski in 944. Several simplified proofs contain gaps. A ne proof presented here combines the ideas published later by G. Mints and M. Aiello, J. van Benthem, G. Bezhanishvili ith a further simplification. The proof strategy is to embed a finite rooted Kripke structure K for S4 into a subspace of the Cantor space hich in turn encodes (, ). This provides an open and continuous map from (, ) onto the topological space corresponding to K. The completeness follos as S4 is complete ith respect to the class of all finite rooted Kripke structures. Introduction The correspondence beteen elementary topology and the modal logic S4 as first established by McKinsey. In [] McKinsey introduced the topological interpretation of S4 here the necessitation connective is interpreted as the topological interior. McKinsey shoed that S4 is complete for the class of all topological spaces. Later more mathematically interesting results ere obtained by McKinsey and Tarski [2], [3]. McKinsey and Tarski shoed that S4 is complete for any dense-in-itself separable metric space. As a consequence, S4 is complete for the real line R, the n-dimensional Euclidean space R n, the Cantor set and the real segment (, ) etc. Recently several attempts ere made to simplify the proof by McKinsey and Tarski. Mints gave a completeness proof of S4 for the Cantor set [4] and a completeness proof of the intuitionistic propositional logic for the real segment (, ) ([5], Chapter 9). Aiello, van Benthem and Bezhanishvili gave a completeness proof of S4 for (, ) ([6], Section 5). Hoever, simplified proofs in [6], Section 5 and [5], Chapter 9 contain gaps. We present here a ne proof, hich combines the ideas in [4], [5] and [6], and provides a further simplification. It goes by () encoding reals in (, ) using a Cantor set B, (2) uninding a finite rooted Kripke structure K for S4 to cover B. Step gives a one-to-one correspondence beteen elements of B (infinite paths in the full binary tree) and real numbers in (, ). Step 2 generates a labeling of the full binary tree by orlds in K and hence establishes a one-to-one correspondence beteen infinite paths in B and infinite sequences of orlds in K. Hence e have a one-to-one correspondence beteen reals in (, ) and infinite sequences of orlds in K. Since K is finite, every infinite sequence of orlds must eventually enter a stable loop hich consists of equivalent orlds ith respect to the frame relation. For each such sequence e pick the label at the stabilization point here the sequence enters the loop. We map each real in (, ) to the label of its corresponding sequence. This provides an open and continuous map from (, ) onto the topological space corresponding to K. The completeness follos as S4 is complete ith respect to the class of finite rooted Kripke structures.
2 Grigori Mints and Ting Zhang We assume basic topology terminology. In particular, e use Int and Cl to denote the interior and closure operators respectively. Definition. (Topological Model) A topological model is an ordered pair M = X, V, here X is a topological space and V is a function assigning a subset of X to each propositional variable. The valuation V is extended to all S4 formulas as follos: V (α β) = V (α) V (β), V (α & β) = V (α) V (β), V ( α) = X \ V (α), V ( α) = Int(V (α)). We say that α is valid in a topological model M and rite M = α if and only if V (α) = X. Definition.2 (Kripke Model) A Kripke frame (for S4) is an ordered pair F = W, R here W is a non-empty set and R is a reflexive and transitive relation on W. The elements in W are called orlds. We say that a orld is an R-successor of a orld if R, and is R-equivalent to (ritten R ) if both R and R. A Kripke frame is rooted if there exists a orld such that any orld in W is an R-successor of. A Kripke model is a tuple M = W, R, V ith W, R a Kripke frame and V a valuation function, hich assigns a subset of orlds in W to every propositional variable. Validity relation = is defined recursively in the standard ay. In particular, (M, ) = α iff (M, ) = α for every such that R. We say that a formula α is valid in M if and only if (M, ) = α for every W. A formula α is valid (ritten = α) if α is valid in every Kripke model. We can think of a Kripke frame as being a topological space by imposing a topology on it. Definition.3 (Kripke Space) Let K = W, R be a Kripke frame. A Kripke space on K is a topological space T = W, O here W is the carrier and O is the collection of all subsets of W closed under R: M O iff ( M and R implies M) for all, W. It is ell-knon that S4 is complete for finite rooted Kripke models [7]. Theorem. For any S4 formula α, S4 α if and only if α is valid in all finite rooted Kripke frames. 2 A correspondence beteen (, ) and finite Kripke structures 2. Binary encoding of real numbers Let Σ = {, }, and let Σ ω be the full infinite binary tree here each node in the tree is identified by a finite path (a finite Σ-ord) from the root Λ to it. We use b and b to denote finite paths and infinite paths
A proof of topological completeness for S4 in (,) 3 respectively. Let C be the standard Cantor set represented by Σ ω, here each element of C is identified ith an infinite path (an infinite Σ-ord). For each b C, b n denotes the prefix of length n, i.e., the finite sequence b n = b()b(2)... b(n). We rite b n b 2 if b n = b 2 n. One can imagine adding the component b() = to account for the root Λ, but e do not do that. (See Figure.) Λ ω ω ω ω.5 Fig.. The full infinite binary tree Let B = C \ ({ ω } Σ ω ), i.e., B is obtained by deleting from C the leftmost path hich corresponds to the ord ω, as ell as paths going right from some point on, hich correspond to sequences ith the infinite tail of s. So for each path b B, b either alays goes left from some point on, or goes both left and right infinitely often. In the former case b ends ith ω and in the latter case b contains infinitely many s as ell as infinitely many s. Formally let B = {b B b = b b 2... b i ω for some i > }, B 2 = B \ B. We vie a sequence in B as a binary encoding of a real number in (, ). A one-to-one correspondence beteen B and (, ) is given by X real(b) = b(i)2 i, i= B(x) = the unique b B such that real(b) = x. The sequences in B represent binary rational numbers in (, ); the sequences in B 2 represent all other real numbers in (, ). For example,.375, in binary., is represented by ω. No it should be clear hy B excludes some binary sequences; sequence ω represents, and numbers represented by sequences of the form b b 2... b n ω can also be represented by sequences of the form b b 2... b n ω. Proposition 2. Let x, y (, ), B(x) (n + ) = b b 2... b n and B(y) (n + ) = b b 2... b n. Then for any z (, ), if x < z < y, then B(z) n = b b 2... b n. Proof. It follos immediately from basic properties of the binary representation.
4 Grigori Mints and Ting Zhang Proposition 2.2 Let x (, ) and B(x) = b b 2... b n ω. Then for any y (, ),. if < x y < 2 (n+2), then B(y) (n + 2) = b b 2... b n, and 2. if < y x < 2 (n+2), then B(y) (n + 2) = b b 2... b n. Proof. (See Figure 2.) Let b = b b 2... b n ω, b = b b 2... b n ω, l = b b 2... b n ω and u = b b 2... b n ω. We kno that B(x) = real(b) = real(b ). Let l = real(l) and u = real(u). Since l + 2 (n+2) = x, for any y such that < x y < 2 (n+2), l < y < x, and so by Proposition 2. B(y) has the prefix b b 2... b n. Similarly, since x + 2 (n+2) = u, for any y such that < y x < 2 (n+2), x < y < l, and so B(y) has the prefix b b 2... b n. Λ b n 2 b n b n ω ω ω ω l y B(x) y u Fig. 2. Proposition 2.2 Proposition 2.3 Let x, y (, ). If B(x) n B(y), then x y < 2 n. Proof. If B(x) n B(y), then obviously x y 2 n. To have x y = 2 n, one of B(x) and B(y) must end ith ω and the other must end ith ω. Since paths ending ith ω has been excluded from B, e have x y < 2 n. Proposition 2.4 Let x (, ) and B(x) (n + 2) = b b 2... b n. If y (, ), y x < 2 (n+2), then B(y) n = b... b n and B(y) b b 2... b n ω. Proof. (See Figure 3.) Let u = b b 2... b n ω, l = b b 2... b n ω, u = b b 2... b n ω, l = b b 2... b n ω, u 2 = b b 2... b n ω, and l 2 = b b 2... b n ω. Let u = real(u), l = real(l), u = real(u ), l =
A proof of topological completeness for S4 in (,) 5 real(l ), u 2 = real(u 2) and l 2 = real(l 2). Obviously, e have l +2 (n+2) = u = l, l+2 (n+2) = u = l 2 and l 2+2 (n+2) = u 2. Since B(x) (n+2) = b b 2... b n, l x < u and so l < y < u 2 as y x < 2 (n+2). Hence by Proposition 2. B(y) n = b... b n and B(y) b b 2... b n ω. ω b n ω ω ω Λ ω ω l ( y) y u (=l) x u(=l 2 ) u 2 Fig. 3. Proposition 2.4 Proposition 2.5 For any x, y (, ) if B(x) = b b 2... b n ω, B(y) m = b b 2... b n m (n+2), then B(y) b b 2... b n m (n+2) ω if and only if x y < 2 m. Proof. Let u = b b 2... b n ω, l = b b 2... b n m (n+2) ω and u = real(u), l = real(l). Obviously, u = x and u l = 2 m. Since B(y) m = b b 2... b n m (n+2), y < u = x. If x y < 2 m, then y > l and so B(y) l. On the other hand, if B(y) l and B(y) m = b b 2... b n m (n+2), then l < y < u and so x y < 2 m. 2.2 Uninding a finite rooted model into the Cantor space Let K = W, R be a finite Kripke model ith root and K be the corresponding Kripke space. In the folloing sections by Kripke model e alays mean a finite rooted one. Definition 2. (Uninding and Labeling) The labeling function W : Σ W is defined recursively as follos. (See Figure 4.). W(Λ) =. 2. Let b Σ be a node in B. Suppose b is already labeled by a orld (i.e., W( b) = ), hile none of its children has yet been labeled. Let,,..., m be all R-successors of. Then
6 Grigori Mints and Ting Zhang W( b i ) = for < i 2m, W( b 2i ) = i for < i m, W( b 2i ) = for i < m. Note that in placing R-successors of at right branches b 2i (i > ), e interleave ith each of its other successors. This is the main distinction from the construction in [6]. (See Figure 4.) n 2 Fig. 4. Uninding and labeling Definition 2.2 (Monotonic Sequences) Let K = W, R be a Kripke model. An infinite sequence b of orlds in W is monotonic (ith respect to R) if Rb(i)b(j) holds for any i < j. We rite W for the set of all monotonic sequences in W ω. By Definition 2. each path in B is labeled by a monotonic sequence in W. We rite W for the induced map from B to W, i.e., W (b) = λn : ω.w(b n) = W(b )W(b 2).... Proposition 2.6 Let W( b) =, W( b) = 2. If 2, then W( b) =. Proof. If W( b) =, then W( b) =. In addition, if W( b) = 2, then R 2. But since 2, and is interleaved ith any other proper successor of during the uninding process, b must be labeled by, that is, W( b) =. Proposition 2.7 Let W( b) =. Then for any W ith R there exist infinitely many i > such that W( b i ) =.
A proof of topological completeness for S4 in (,) 7 Proof. Let,,..., m be all R-successors of. By Definition 2. W( b 2i ) = i for < i m and W( b 2m ) =. By our definition, e have for all k W( b 2mk+(2i ) ) = i for < i m. Proposition 2.8 Let W( b) =. If W( b) =, then for any i, W( b i ) =. Proof. Note that b gets labeled only after b has been labeled. By Definition 2., W( b) =. Repeating this argument e have W( b i ) = for any i. Definition 2.3 (Stabilization Point) We say a point i is a stabilization point for a monotonic sequence b if b(i) R b(j) for any j > i. If K is a finite model, each sequence in W must eventually enter a stable loop consisting of R- equivalent orlds. (Note that the loop may consist of a single orld.) We define function λ : B N by λ(b) = µn[n & ( i, j n RW (b)(i)w (b)(j))]. In other ords, the function λ returns the non-root R-stabilization point of W (b) for each b B. Definition 2.4 We define a map δ : B N as follos: ( δ (b) if b B δ(b) = δ 2(b) if b B 2 here δ : B N is defined by and δ 2 : B 2 N is defined by δ (b) = max(, n), if b = b b 2... b n ω, δ 2(b) = µn(n > λ(b) & b(n) = & b(n ) = ). The map δ ill serve as the modulus of continuity for the map π : B W introduced belo. Definition 2.5 (Selection Function) We define a selection function ρ : B N and a map π : B W as follos: (See Figure 5.) ( δ (b) if b B, ρ(b) = λ(b) if b B 2. π(b) = W (b)(ρ(b)). For notation simplicity e identify function f : B X ith the corresponding function B f : (, ) X. For example, ρ(x) (x (, )) should be understood as ρ(b(x)). In particular, π(x) = W (B(x))(ρ(B(x))).
8 Grigori Mints and Ting Zhang ρ(b) ω ρ(b)=λ(b) x y Fig. 5. The selection function ρ: Case b B (left) and Case b B 2 (right) 3 Proof of completeness Lemma 3. If b B ith b(n + ) = (n > ), then ρ(b) n. Proof. Suppose that b = b... b m ω. Since b(n + ) = and n >, e have m > and m = ρ(b). It follos immediately that n m as b has the prefix b... b n. Lemma 3.2 If b B 2, then either ρ(b) = λ(b) = or b(λ(b)) = b(ρ(b)) =. Proof. For b B 2 either W (b) stabilizes at the root, or W (b) stabilizes at point n for n >. In the former case, ρ(b) = λ(b) =. In the latter case, e must have b(n) =. Otherise W (b) stabilizes at point n as W(b)(n ) = W(b)(n). So b(λ(b)) = b(ρ(b)) =. Lemma 3.3 Let b, b 2 B ith ρ(b ) = n, ρ(b 2) = n 2. If n n 2 and b n b 2, then Rπ(b )π(b 2). Proof. Since n n 2 and b n b 2, the node b 2 n 2 (labeled by W (b 2)(n 2)) is in the subtree ith root b n (labeled by W (b )(n )). So RW (b )(n )W (b 2)(n 2), that is, Rπ(b )π(b 2). Lemma 3.4 Let b B ith ρ(b ) = n, b 2 B 2. If b n b 2, then Rπ(b )π(b 2). Proof. Let ρ(b 2) = λ(b 2) = n 2. If n n 2, then Rπ(b )π(b 2) by Lemma 3.3. Suppose that n > n 2. Since b n b 2 and n 2 is the stabilization point of W (b 2), W (b )(n ) is in the final stabilization loop here W (b 2)(n 2) belongs. So R (n ) 2(n 2), i.e., Rπ(b )π(b 2).
A proof of topological completeness for S4 in (,) 9 Lemma 3.5 For any x, y (, ), if y x < 2 (δ(x)+2), then Rπ(x)π(y). Proof. Consider all possible cases.. Case B(x) B. (See Figure 6.) If B(x) = ω, then π(x) =, the root of K, and trivially Rπ(x)π(y) for any y (, ). Suppose that B(x) = b b 2... b n ω for n. Then ρ(x) = δ(x) = δ (x) = n, y x < 2 (n+2) and B(x)(n+) =. a) Case y < x. By Proposition 2.2, B(y) has b b 2... b n as a prefix. So B(x) n B(y) and B(y)(n + 2) =. If B(y) B, then by Lemma 3. ρ(y) = δ (y) n + > n = ρ(x). By Lemma 3.3 Rπ(x)π(y). If B(y) B 2, then ρ(y) = λ(y). Since B(x) n B(y), by Lemma 3.4 Rπ(x)π(y). b) Case y > x. By Proposition 2.2, B(y) has b b 2... b n as a prefix. So B(x) n+2 B(y) and B(y)(n + ) =. If B(y) B, then again by Lemma 3. ρ(y) = δ (y) n and by Lemma 3.3 Rπ(x)π(y). If B(y) B 2, then ρ(y) = λ(y). Since B(x) n B(y), by Lemma 3.4 Rπ(x)π(y). ω ω ω x 2 (δ(x)+2) y x y x+2 (δ(x)+2) ω Fig. 6. Lemma 3.5 Case B(x) B 2. Case B(x) B 2. (See Figures 7, 8.) Let ρ(x) = λ(x) = m and δ(x) = n. If m =, then π(x) =, the root of K, and so Rπ(x)π(y) for any y (, ). Suppose that m >. By Lemma 3.2 B(x)(m) =. By Definition 2.4 n = δ(x) > m = λ(x), n m + 2. Assume that B(x) n = b b 2... b m b m+... b n 2 Since y x < 2 (n+2), by Proposition 2.4 B(y) (n 2) = b b 2... b m b m+... b n 2 and B(y) b b 2... b n 2 ω
Grigori Mints and Ting Zhang a) Case B(y) B. (See Figure 7.) It immediately follos from the above condition that for any i n 2 B(y) b b 2... b i ω Then n is the least value k for hich it is possible to have B(y) = b b 2... b k ω By Proposition 3. ρ(y) n > m. Since B(y) m B(x) and ρ(x) = m, by Lemma 3.3 Rπ(x)π(y). ω π(x) ω a y x y a+2 (δ(x)+2) ω ω Fig. 7. Lemma 3.5 Case B(x) B 2 & B(y) B b) Case B(y) B 2. (See Figure 8.) As n 2 m, B(y) m B(x). Since ρ(x) = m, by Lemma 3.4 Rπ(x)π(y). Lemma 3.6 For any x (, ), ɛ >, W ith Rπ(x), there exists y (, ) such that y x < ɛ and π(y) =. Proof.. Case B(x) B 2. (See Figure 9.) Let m = λ(x) and take n > m such that 2 n < ɛ. Assume that
A proof of topological completeness for S4 in (,) ω π(x) a y x y a+2 (δ(x)+2) ω Fig. 8. Lemma 3.5 Case B(x) B 2 & B(y) B 2 B(x) n = b b 2... b m... b n, W(b... b m) =, W(b... b m... b n) = 2 By the assumption π(x) = and, 2 are R-equivalent. By Proposition 2.7 for any R-successor of (and hence of 2) there exists i such that Let W(b... b m... b n i ) = b = b... b m... b n i ω Then b B and π(b) =. By Proposition 2.3, for any b, b 2, B, if b n b 2, then B (b ) B (b 2) < 2 n. Take y = B (b), then e have y x < 2 n < ɛ and π(y) =. 2. Case B(x) B. (See Figures,.) Suppose that B(x) = b b 2... b n ω, W(b b 2... b n) =, W(b b 2... b n) = 2. We have ρ(x) = δ (x) = n, π(x) =. a) Case = 2. (See Figure.) Let W be an R-successor of 2. By Proposition 2.7 there exists m > n + such that 2 m < ɛ and W(b... b n m (n+) ) =. Let
2 Grigori Mints and Ting Zhang ω 2 i 2 ω a y x a+2 n ω Fig. 9. Lemma 3.6 Case B(x) B 2 b = b... b n m (n+) ω Then b B and π(b) =. No let y = B (b), so π(y) =. Since B(x) m B(y), by Proposition 2.3, y x < 2 m < ɛ as desired. b) Case 2. (See Figure.) Let W ith R. By Proposition 2.6 W(b b 2... b n) = By Proposition 2.8 e can take m > n + 2 such that W(b b 2... b n m (n+2) ) = By Proposition 2.7 there exists k > m such that Let W(b b 2... b n m (n+2) k m ) = b = b b 2... b n m (n+2) k m ω Then b B and π(b) =. No let y = B (b), so π(y) =. By Proposition 2.5, y x < 2 m < ɛ as desired.
A proof of topological completeness for S4 in (,) 3 m (n+) 2 2 2 2 ω ω ω x y x+2 m Fig.. Lemma 3.6 Case B(x) B & = 2 Theorem 3. The function π is an open and continuous from B onto the Kripke space K. Proof.. Continuity. Let W W be an open set of the Kripke space K (i.e., W is closed under R). For any W, let x π (), i.e., π(x) =. Take a set O x = {y x y < 2 (δ(x)+2) }. Obviously O x is an open subset of (, ). By Lemma 3.5 all orlds in π(o x) are R-successors of. Since W and W is closed under R, e have π(o x) W. Hence π is continuous. 2. Openness. Let O x be the collection of sets O x,i = {y x y < 2 (i+δ(x)+2) } for i. Clearly S x Ox is a base of the standard topology on (, ). By Lemma 3.5 for any π(o x,i) e have Rπ(x). And by Lemma 3.6 for any ith Rπ(x), there exists y O x,i such that π(y) =, that is, π(o x,i). Hence π(o x,i) = { W Rπ(x)}, hich is obviously closed under R. Hence π is an open map. Lemma 3.7 Let X, X 2 be to topological spaces and f : X X 2 a continuous and open map. Let V 2 be a valuation for topological semantics on X 2 and define for each propositional variable p. Then for any S4-formula α. V (p) = f (V 2(p)) () V (α) = f (V 2(α))
4 Grigori Mints and Ting Zhang 2 2 m (n+2) 2 ω k m ω ω ω x 2 m y x Fig.. Lemma 3.6 Case B(x) B & 2 Proof. The proof uses induction. The base case and induction steps for connectives, &, are straightforard. No suppose α = β. By induction hypothesis, It follos from openness and continuity that Hence e have V (β) = f (V 2(β)). Int(f (V 2(β))) = f (Int(V 2(β))). V (α) = V ( β) = Int(V (β)) = Int(f (V 2(β))) = f (Int(V 2(β))) = f (V 2( β)) = f (V 2(α)). Lemma 3.8 Let X, X 2 be to topological spaces and f : X X 2 a continuous and open map. Let V 2 be a valuation for topological semantics on X 2 and define V by the equation (). Then for any S4-formula α, X 2, V 2 = α implies X, V = α. Moreover if f is onto, then X 2, V 2 = α iff X, V = α.
A proof of topological completeness for S4 in (,) 5 Proof. Suppose X 2, V 2 = α, that is, V 2(α) = X 2. By Lemma 3.7 V (α) = f (V 2(α)), and so V (α) = X as required. No suppose that f is onto and X, V = α, but X 2, V 2 = α, i.e., V 2(α) X 2. Since f is onto and V (α) = f (V 2(α)), e have V (α) X, that is, X, V = α, a contradiction. Theorem 3.2 S4 is complete for the standard topology on (, ). Proof. It suffices to sho that every non-theorem of S4 can be refuted on (, ). Let α be such an nontheorem. We need to find a valuation V such that V (α) (, ). By Theorem. there exists a finite rooted Kripke model K = X, V such that K = α. By Theorem 3., e have a continuous and open map π from (, ) onto K. Let V be the valuation on (, ) such that V (p) = π (V (p)) for every propositional variable p. By Lemma 3.8 V (β) = X if and only if V (β) = (, ) for any S4- formula β. In particular since V (α) X, V (α) (, ). It follos that S4 is complete for (, ). References. J. C. C. McKinsey. A solution of the decision problem for the Leis systems S2 and S4, ith an application to topology. The Journal of Symbolic Logic, 6(4):7 34, 94. 2. J. C. C. McKinsey and Alfred Tarski. The algebra of topology. The Annals of Mathematics, 45():4 9, January 944. 3. Helena Rasioa and Roman Sikorski. The mathematics of metamathematics. Polish Scientific Publishers, Warsa, 968. 4. Grigori Mints. A completeness proof for propositional S4 in Cantor space. In Ea Orloska, editor, Logic at ork: Essays dedicated to the memory of Helena Rasioa, pages 79 88. Heidelberg: Physica- Verlag, 999. 5. Grigori Mints. A short introduction to intuitionistic logic. Kluer Academic-Plenum Publishers, Ne York, 2. 6. Marco Aiello, Johan van Benthem, and Guram Bezhanishvili. Reasoning about space: the modal ay. Technical Report PP-2-8, ILLC, Amsterdam, 2. 7. Saul Kripke. Semantical analysis of modal logic I, normal propositional calculi. Zeitschrift fur Mathematische Logik und Grundlagen der Mathematik, 9:67 96, 963.